# Using Borel Cantelli lemma to show that the set of convergence of non degenerate independent random variables has measure zero.

I'm trying the following:

If $$X_1,\dots, X_n,\dots,$$ are non degenerate independent and identically distributed random variables; then

$$\begin{equation*} \mathbb{P}\left(X_n \text{ converge }\right)=0 \end{equation*}$$ I tried a lot this exercise; but there is something that I'm not using. Firstly, I wrote the set of convergence as:

$$\begin{equation} \bigcap_{k=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{m=n+1}^{\infty}\lbrace \mid X_n-X_m\mid\le \frac{1}{k}\rbrace \end{equation}$$

At this point, I can't see how to use Borel Cantelli lemma, even I can't see where I have to use the hypothesis of independence. I tried to write the set $${\mid X_n-X_m\mid}$$ and making a change in the index to obtain independent set, but it's not working.

By the Kolmogorov One Zero law one can show that the set of convergence of $$X_n$$ has probability either 0 or 1. Using this, I tried to obtain a contradiction if I suppose that $$\mathbb{P}\left(X_n \text{ converge }\right)=1$$.

The thing that I can't see how to use is the hypothesis that the variables are non degenerate.

I hope that you can provide a hint or something!

Thanks

EDIT: The set where the sequence $$X_n$$ converge is the same where $$\limsup X_n$$ and $$\liminf X_n$$ are equal. Since $$\limsup X_n$$ is measurable with respect to tail sigma-algebra, it has to be constant a.s. Thus, if $$X_n$$ converge a.s., $$X_n$$ is constant eventually a.s.; which is a contradiction. Is this "intuitive" argument valid?

• The hypothesis that the variables are non-degenerate is to account for cases like $X_i = 1$ almost-surely. Sep 27, 2020 at 20:51
• I suppose that I have to use the fact that for every constant c, $P(X=c)<1$ to obtain estimates, but How I can use it to obtain such estimates? Sep 27, 2020 at 21:56

Let $$\alpha = P(X_1 = X_2)$$.
If $$\alpha < 1$$, by the dominated convergence theorem, there exists $$\epsilon>0$$ such that for any $$m \ne n$$, $$P(|X_m - X_n| > \epsilon) = P(|X_1 - X_2| > \epsilon) > 0$$. So $$\sum_{n} P(|X_{2n} - X_{2n+1}| > \epsilon) = \infty$$. Now use the Borel-Cantelli Lemma.
If $$\alpha = 1$$, suppose for a contradiction that $$X_1$$ is not degenerate. Then there exists $$c$$ such that $$P(X_1 > c)$$ and $$P(X_1 \le c)$$ are both non-zero. But then $$P(X_1 \ne X_2) \ge 2 P(X_1 > c) P(X_1 \le c) > 0$$.
• Thanks! I was thinking that non-degeneracy is useful to obtain estimates to show the divergence of the sum; but I can't see the relation with the dominated convergence theorem. Are you using the indicator functions $1_{\mid X_m-X_n\mid>\epsilon}$? Sep 28, 2020 at 18:00